Proof

We begin by demonstrating that \(\text{cos} x\) is continuous at every real number. To do this, we must show that \(\underset{x \rightarrow a}{\text{lim}} \text{cos} x = \text{cos} a\) for all values of a.

\(\begin{aligned} \underset{x \rightarrow a}{\text{lim}} \text{cos} x & = \underset{x \rightarrow a}{\text{lim}} \text{cos} \left(\right. \left(\right. x - a \left.\right) + a \left.\right) & & & \text{rewrite} x = x - a + a \\ & = \underset{x \rightarrow a}{\text{lim}} \left(\right. \text{cos} \left(\right. x - a \left.\right) \text{cos} a - \text{sin} \left(\right. x - a \left.\right) \text{sin} a \left.\right) & & & \text{apply the identity for the cosine of the sum of two angles} \\ & = \text{cos} \left(\right. \underset{x \rightarrow a}{\text{lim}} \left(\right. x - a \left.\right) \left.\right) \text{cos} a - \text{sin} \left(\right. \underset{x \rightarrow a}{\text{lim}} \left(\right. x - a \left.\right) \left.\right) \text{sin} a & & & \underset{x \rightarrow a}{\text{lim}} \left(\right. x - a \left.\right) = 0 , \text{and} \text{sin} x \text{and} \text{cos} x \text{are continuous at 0} \\ & = \text{cos} \left(\right. 0 \left.\right) \text{cos} a - \text{sin} \left(\right. 0 \left.\right) \text{sin} a & & & \text{evaluate cos}(\text{0})\text{ and sin}(\text{0})\text{ and simplify} \\ & = 1 \cdot \text{cos} a - 0 \cdot \text{sin} a = \text{cos} a . \end{aligned}\)

The proof that \(\text{sin} x\) is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of \(\text{sin} x\) and \(\text{cos} x ,\) their continuity follows from the quotient limit law.

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As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.